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(II) Two cords support a chandelier in the manner shown in Fig. 9–4 except that the upper cord makes an angle of 45° with the ceiling. If the cords can sustain a force of 1660 N without breaking, what is the maximum chandelier weight that can be supported?



Short Answer

Expert verified

The maximum chandelier weight that can be supported is \(W = 1173.3\;{\rm{N}}\).

Step by step solution

01

Given data

The angle is\(\theta = 45^\circ \).

The force sustained by the cord is \(F = 1660\;{\rm{N}}\).

02

Understanding Newton’s second law

Newton’s second law describes the variation that a force generates in the movement of a body. In this question, Newton’s second law for the junction will be used in order to determine the weights in the x and y directions.

03

Free body diagram and calculation of forces in the x-direction

The following is the free body diagram.

The relation of the forces in the x-direction can be written as:

\(\begin{array}{c}\sum {F_{\rm{x}}} = 0\\\left[ {{F_{\rm{B}}} - {F_{\rm{A}}}\cos \theta } \right] = 0\\{F_{\rm{B}}} = {F_{\rm{A}}}\cos \theta \end{array}\)

Here, \({F_{\rm{A}}}\)and \({F_{\rm{B}}}\) are the forces acting on the chandelier.

In the above relation, it can be seen that force \({F_{\rm{A}}}\) is greater than force \({F_{\rm{B}}}\). Hence, the maximum force will be considered as \({F_{\rm{A}}} = F = 1660\;{\rm{N}}\).

04

Calculation of forces in the y-direction

The relation of the forces in the y-direction can be written as:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\\left[ {W - {F_{\rm{A}}}\sin \theta } \right] = 0\\W = {F_{\rm{A}}}\sin \theta \end{array}\)

On plugging the values in the above relation, you get:

\(\begin{array}{l}W = \left( {1660\;{\rm{N}}} \right)\sin 45^\circ \\W = 1173.3\;{\rm{N}}\end{array}\)

Thus, \(W = 1173.3\;{\rm{N}}\) is the required weight.

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Most popular questions from this chapter

(II) Find the tension in the two cords shown in Fig. 9–52. Neglect the mass of the cords, and assume that the angle is 33°, and the mass m is 190 kg.

A ground retaining wall is shown in Fig. 9–36a. The ground, particularly when wet, can exert a significant force F on the wall. (a) What force produces the torque to keep the wall upright? (b) Explain why the retaining wall in Fig. 9–36b would be much less likely to overturn than that in Fig. 9–36a.

Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center and a very light board that is 3.2 m long (Fig. 9–57). Two playmates are already on either end. Boy A has a mass of 45 kg, and boy B has a mass of 35 kg. Where should girl C, whose mass is 25 kg, place herself to balance the seesaw?

As you increase the force that you apply while pulling on a rope, which of the following is not affected?

(a) The stress on the rope

(b) The strain on the rope

(c) The Young’s modulus of the rope

(d) All of the above

(e) None of the above

A heavy ball suspended by a cable is pulled to the side by a horizontal force \(\vec F\), as shown in Fig. 9–43. If angle \(\theta \) is small, the magnitude of the force F can be less than the weight of the ball because

(a) the force holds up only part of the ball’s weight.

(b) even though the ball is stationary, it is not really in equilibrium.

(c) \(\vec F\) is equal to only the x component of the tension in the cable.

(d) the original statement is not true. To move the ball, \(\vec F\) must be at least equal to the ball’s weight.

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